Problem: $h(x)=-\dfrac{27}{\sqrt[{3}]{x-6}}$. On which intervals is the graph of $h$ concave up? Choose 1 answer: Choose 1 answer: (Choice A) A $(\sqrt[3]{6},\infty)$ only (Choice B) B $(-\infty,\sqrt[3]{6})$ only (Choice C) C $(6,\infty)$ only (Choice D) D $(-\infty,6)$ only
We can analyze the intervals where $h$ is concave up/down by looking for the intervals where its second derivative $h''$ is positive/negative. This analysis is very similar to finding increasing/decreasing intervals, only instead of analyzing $h'$, we are analyzing $h''$. The second derivative of $h$ is $h''(x)=-\dfrac{12}{\sqrt[3]{(x-6)^7}}$. $h''$ is never equal to $0$. $h''$ is undefined for $x=6$. Therefore, our only point of interest is $x=6$. Our point of interest divides the domain of $h$ (which is all numbers except for $6$ ) into two intervals: $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $(-\infty,6)$ $(6,\infty)$ Let's evaluate $h''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h''(x)$ Verdict $(-\infty,6)$ $x=5$ $h''(5)=12>0$ $h$ is concave up $\cup$ $(6,\infty)$ $x=7$ $h''(7)=-12<0$ $h$ is concave down $\cap$ In conclusion, the graph of $h$ is concave up over the interval $(-\infty,6)$ only.